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• Dec 3rd 2006, 07:06 PM
Tascja
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The surface area of a cube is changing at a rate of 8 cm^2/s. How fast is the volume changing when the surface area is 60cm^2?

How would i find the length and the rate of change of the length??
• Dec 3rd 2006, 08:46 PM
CaptainBlack
Quote:

Originally Posted by Tascja
The surface area of a cube is changing at a rate of 8 cm^2/s. How fast is the volume changing when the surface area is 60cm^2?

How would i find the length and the rate of change of the length??

Volume of cube, of side length $\displaystyle l$ is:

$\displaystyle V=l^3$.

Surface area of the same cube is:

$\displaystyle A=6 \, l^2$

So:

$\displaystyle \frac{dV}{dt}=3\,l^2\, \frac{dl}{dt}$

and:

$\displaystyle \frac{dA}{dt}=12\,l\, \frac{dl}{dt}$.

Now when the area is $\displaystyle 60\ \rm{cm^2}$ $\displaystyle l^2=10\ \rm{cm^2}$, so:

$\displaystyle \frac{dV}{dt}=30\, \frac{dl}{dt}$.

Also:

$\displaystyle \frac{dA}{dt}=12 \sqrt{10}\, \frac{dl}{dt}=8\,\rm{cm^2/s}$,

so:

$\displaystyle \frac{dl}{dt}=\frac{2}{3\, \sqrt{10}}$,

hence:

$\displaystyle \frac{dV}{dt}=\frac{30\times 2}{3\,\sqrt(10}=2\,\sqrt{10}$

RonL